When an electron falls from a higher energy level to a lower energy level in the hydrogen atom, it emits a photon with a specific wavelength. This wavelength can be used to determine the energy difference between the two energy levels involved in the transition. The energy of a photon is given by:
E = hc/λ
where E is the energy of the photon, h is Planck's constant, c is the speed of light, and λ is the wavelength of the photon.
We can use the energy of the photon to determine the energy difference between the initial and final energy levels of the electron in the hydrogen atom. This energy difference is given by:
ΔE = E_initial - E_final
where ΔE is the energy difference, E_initial is the initial energy level, and E_final is the final energy level.
The energy difference can be calculated by rearranging the formula for the energy of the photon as follows:
ΔE = hc/λ
Substituting the given values, we get:
ΔE = (6.626 x 10^-34 J s) x (3.00 x 10^8 m/s) / (434.1 x 10^-9 m)
ΔE = 4.564 x 10^-19 J
The energy difference between the two energy levels is equal to the initial energy level minus the final energy level. The final energy level is n2, and we need to find the initial energy level. Using the formula for the energy levels of the hydrogen atom:
E_n = -13.6 eV / n^2
where E_n is the energy of the nth energy level in electron volts (eV), and n is the principal quantum number.
We can set up the following equation to solve for the initial energy level:
E_initial - E_final = (-13.6 eV / n_initial^2) - (-13.6 eV / n2^2)
Substituting the values of ΔE and n2, we get:
4.564 x 10^-19 J = (-13.6 eV / n_initial^2) - (-13.6 eV / 2^2)
Simplifying this equation and solving for n_initial, we get:
n_initial = 3
Therefore, the initial energy level of the electron in the hydrogen atom was n = 3.